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Passive Elements and Phasor Diagrams Resistor v Ri V = RI R i + v - v i I Inductor i di dt V = jwL I vL L + V v - v i V I Capacitor + v i dv dt I = jwC V C i v iC - I V ELEC 371 Three-phase systems 1 Industrial Power Systems Salvador Acevedo Ideal Transformer i1 + v1 - v1 i 2 N1 a v 2 i1 N2 V1 I2 N1 a V2 I1 N2 i2 + v2 - N1:N2 Transformer feeding load: I1 I2 + V1 - + V2 - Z V2 = V1/a I2 = V2/Z I1= I2/a V2 Assuming a RL load connected to secondary and ideal source to primary V1 I1 I2 ELEC 371 Three-phase systems 2 Industrial Power Systems Salvador Acevedo Two Winding Transformer Model The linear equivalent model of a real transformer consists of an ideal transformer and some passive elements i1 i2 + + v1 v2 - ELEC 371 Three-phase systems 3 N1:N2 - Industrial Power Systems Salvador Acevedo AC Generators and Motors AC synchronous generator Single-phase equivalent AC synchronous motor Single-phase equivalent AC induction motor (rarely used as generator) ELEC 371 Three-phase systems 4 Industrial Power Systems Salvador Acevedo Steady-state Solution In sinusoidal steady-state a circuit may be solved using phasors I R vs + VS - jwL i 0 VS = R I p 2p Ix=I cos + jwL I VS = (R + jwL) I VS VS = (R + jX) I VS = Z I VS Z Vmax 0 I Z I I = Iy=I sin Rectangular form I I max I Polar form I = Ix + j Iy = Imax From rectangular form to polar form: I = Ix 2 + Iy 2 Iy Ix tan 1 Magnitude Angle or phase From polar form to rectangular form: Ix Icos Real part Iy Isin Reactive or imaginary part ELEC 371 Three-phase systems 5 Industrial Power Systems Salvador Acevedo Single-phase Power Definitions i(t) = Im sin (wt+ i) amps + - v(t) = Vm sin(wt+ v) volts Load: any R,L,C combination w: angular frequency in rad/sec f: frequency in Hz w=2pf Instantaneous power p( t ) v ( t ) i ( t ) Vm sin wt v p( t ) I m ) sin( wt i ) 1 V I cos(v i ) cos( 2 wt v 2 m m i Average Power (or REAL POWER) 1 P T T p( t ) dt 0 1 V I cos VrmsI rms cos 2 m m Apparent Power S = VrmsI rms Power Factor REAL POWER P pf APPARENT POWER S For this circuit, the power factor is pf = Vrms I rms cos cos Vrms I rms ELEC 371 Three-phase systems 6 Industrial Power Systems Salvador Acevedo Power Triangle S Ssin =Q P=Scos Real Power P = S cos = V I cos Reactive Power Q = S sin V I sin vars watts Complex Power S = S P + j Q S = VI cos + jVIsin If = v - i and assuming a reference then = -i therefore V Icos( ) - v = 0 S = V Icos(-i ) + jIsin(-i ) S= i jIsin(i ) S V I* The magnitude is called Apparent Power: S = VI volt - amperes (VA) ELEC 371 Three-phase systems 7 Industrial Power Systems Salvador Acevedo Power Consumption by Passive Elements Impedance: Z = R + jX = Z Resistive Load Z = R = R 0o P = VI cos0o = VI = I 2 R = V 2 / R Q = VI sin 0o = 0 watts vars A resistor absorbs P Purely Inductive Load Z = jwL = jX L X L 90o P VI cos(90o ) = 0 watts Q = VI sin(90o ) = VI = I 2 X L = V 2 / X L var s An inductor absorbs Q Purely Capacitive Load Z= 1 = -jX C X C 90o jwC P VI cos(-90o ) = 0 watts Q = VI sin(-90o ) = -VI = -I 2 X L = -V 2 / X L var s A capacitor absorbs negative Q. It supplies Q. ELEC 371 Three-phase systems 8 Industrial Power Systems Salvador Acevedo Advantages of Three-phase Systems Creation of the three-phase induction motor Starting torque Three-phase induction motor Single-phase induction motor yes no needs auxiliary starting circuitry Steady state torque Constant Oscillating causing vibration Efficient transmission of electric power 3 times the power than a single-phase circuit by adding an extra cable i va + v - Single-phase Load vb vc p = vi ia ib Three-phase Load ic p = va ia + vb ib + vc ic Savings in magnetic core when constructing Transformers Generators ELEC 371 Three-phase systems 9 Industrial Power Systems Salvador Acevedo Three-phase Voltages va vb vc va(t) = Vm sin wt volts vb(t) = Vm sin (wt - 2p/3) = Vm sin (wt - 120) volts vc(t) = Vm sin (wt - 4p/3) = Vm sin (wt - 240) or volts vb(t) = Vm sin (wt + 2p/3) = Vm sin (wt + 120) w=2p f w: angular frequency in rad/sec volts f : frequency in Hertz Vc 120 120 Va 120 ELEC 371 Three-phase systems 10 Vb Industrial Power Systems Salvador Acevedo Star Connection (Y) Y-connected Voltage Source a - n c + + Van - Vbn + Vcn b Line - to - neutral voltages Van, Vbn, Vcn. (phase voltages for Y - connection) same magnitude: V P VP Van Vbn Vcn Line - to - line voltages Vab, Vbc, Vca same magnitude: VLL Vab = Van - Vbn VLL = ELEC 371 Three-phase systems 11 3 VP Industrial Power Systems Salvador Acevedo Delta Connection (D ) D-connected Voltage Source a Vca + Vab - + c Vbc + b Line - to - line voltages Vab, Vbc, Vca. (phase voltages for D - connection) same magnitude: VLL VP Phase currents Iab, Ibc, Ica. same magnitude: I P Line currents Ia, Ib, Ic. same magnitude: I L IL = 3 IP ELEC 371 Three-phase systems 12 Industrial Power Systems Salvador Acevedo Y-connected Load Ia a - n c + ia + Van - Za Vbn + Vcn n' b Zc Zb Ib Ic ia ia Balanced case: Za = Zb = Zc = Z Ia + Ib + Ic = 0 Ib = Ia -120 Ic = Ia - 240 ELEC 371 Three-phase systems 13 Industrial Power Systems Salvador Acevedo D-connected Load Ia a - n c + ia + Van - Vcn Zca Vbn + b Zab Zbc Ib Ic ia ia ELEC 371 Three-phase systems 14 Industrial Power Systems Salvador Acevedo Y-D Equivalence Za Zca Zab n' Zc Zb Zbc Balanced case: Za = Zb = Zc = Zy Z D = 3Zy Zab = Zbc = Zca = Z D = 3Zy ELEC 371 Three-phase systems 15 Industrial Power Systems Salvador Acevedo Power in Three-phase Circuits Three - phase voltages and currents: va Vm sin wt v ia I m sin wt i vc Vm sin wt v 240 ic I m sin wt i 240 vb Vm sin wt v 120 ib I m sin wt v 120 The three - phase instantaneous power is: p(t ) p3 va ia vb ib vc ic sin wt v sin wt i sin wt v 120 sin wt v 120 p3 Vm I m sin wt v 240 sin wt i 240 This expression can easily be reduced to: p3 23 Vm I m cosv i Since the instantaneous power does not change with the time, its average value equals its intantaneous value: P3 p3 P3 3VP I P cos where: VP ELEC 371 Three-phase systems 16 Vm 2 IP Im 2 v i Industrial Power Systems Salvador Acevedo Three-phase Power In a Y - connection VLL 3 VP IL IP V P3 3VP I P cos 3 LL I L cos 3 VLL I L cos 3 In a D - connection VLL VP IL 3 IP I P3 3VP I P cos 3VLL L cos 3 VLL I L cos 3 Regardless of the connection (for balanced systems), the average power (real power) is : P3 3 VLL I L cos watts Similarly, reactive power and apparent power expressions are: Q 3 3 VLL I L sin vars S3 3 VLL I L VA ELEC 371 Three-phase systems 17 Industrial Power Systems Salvador Acevedo Three-phase Transformers Use of one three-phase transformer unit Use of 3 single phase transformers to form a “Transformer Bank ELEC 371 Three-phase systems 18 Industrial Power Systems Salvador Acevedo Physical Overview ELEC 371 Three-phase systems 19 Industrial Power Systems Salvador Acevedo N1 : N2 A single-phase transformer Example:10 KVA, 38.1KV/3.81KV + v2 - + v1 - a=N1/N2=V1/V2 Three-phase Transformers Connections Y-Y; DD; Y- D; D -Y Primary terminals Bank of 3 single-phase transformers Secondary terminals The Bank Transformation Ratio is defined as: primary voltage a' line to line line to line secondary voltage ELEC 371 Three-phase systems 20 Industrial Power Systems Salvador Acevedo Y-Y connection A B C N N1 N2 + v1 - + v2 - N1 N1 N2 N2 n a b c Ratings for Y-Y bank using 3 single-phase transformers: 3x10KVA = 30 KVA 66 KV / 6.6 KV ELEC 371 Three-phase systems 21 Industrial Power Systems Salvador Acevedo DD connection A N1 N2 + v1 - + v2 - a B C N1 N1 N2 N2 b c Ratings for DD bank: 30 KVA; 38.1 KV / 3.81 KV ELEC 371 Three-phase systems 22 Industrial Power Systems Salvador Acevedo Y- D connection A B C N N1 N2 + v1 - + v2 - a N1 N1 N2 N2 b c Ratings for Y-D bank: 30 KVA; 66 KV / 3.81 KV ELEC 371 Three-phase systems 23 Industrial Power Systems Salvador Acevedo D -Y connection A N1 N2 + v1 - + v2 - B C N1 N1 N2 N2 n a b c Ratings for D-Y bank: 30 KVA; 38.1 KV / 6.6 KV ELEC 371 Three-phase systems 24 Industrial Power Systems Salvador Acevedo Per unit modelling Power lines operate at kilovolts (KV) and kilowatts (KW) or megawatts (MW) To represent a voltage as a percent of a reference value, we first define this BASE VALUE. Example: Base voltage: Vbase = 120 KV Circuit voltage Percent of base value Per unit value 108 KV 90% 0.9 120 KV 100% 1.0 126 KV 105% 1.05 60 KV 50% 0.5 per unit quantity = Voltage_1= actual quantity base quantity 108 0.9 p.u. 120 ** The percent value and the per unit value help the analyzer visualize how close the operating conditions are to their nominal values. ELEC 371 Three-phase systems 25 Industrial Power Systems Salvador Acevedo Defining bases 4 quantities are needed to model a network in per unit system: V: I: S: Z: voltage current power impedance VBASE IBASE SBASE ZBASE Vactual Vbase Sactual S base I actual I base Z actual Z base Vpu I pu S pu Z pu Given two bases, the other two quantities are easily determined. If base voltage and base power are known: Vbase 100 KV, S base 100 MVA then, base current and base impedance are: S base 100,000,000 I base = I base 1000 A. Vbase 100,000 Vbase 100,000 Z base = Z base 100 I base 1000 Another way to express base impedance is: Z base Vbase Vbase = I base S base Vbase V base 2 S base Real power base and reactive power base are: Pbase = S base = 100 MW Q base = S base = 100 MVAR ELEC 371 Three-phase systems 26 Industrial Power Systems Salvador Acevedo Three phase bases In three-phase systems it is common to have data for the three-phase power and the line-to-line voltage. Sbase -3 3S base1 Vbase LL 3Vbase LN The base current and impedance for the three - phase case are : I base S base3 3 Vbase LL 3 S base3 3Vbase LL 2 Z base Vbase LL 2 Vbase LL 3 S base3 S base3 3 In per unit, line - to - neutral voltage = line - to - line voltage VLN(pu) = VLL(pu) why? With p.u. calculations, three-phase values of voltage, current and power can be used without undue anxiety about the result being a factor of 3 incorrect !!! ELEC 371 Three-phase systems 27 Industrial Power Systems Salvador Acevedo Example The following data apply to a three-phase case: Sbase=300 MVA (three-phase power) Vbase=100 KV (line-to-line voltage) a b c Three-phase load 270 MW 100 KV pf=0.8 Using the per unit method: 270 P= 0.9 p.u. 300 V = 1.0 p.u. Normally, we' d say: P = 3 VL I L cos = 3 VL I L pf I= P 3 VL pf 270x106 3 (100x103 ) (0.8) 1948.5 A. Single-phase equivalent: I=1.125 p.u. P = V I pf then P 0.9 I= 1125 . p. u. V pf (10 . )(0.8) + V=1 p.u. - This current is 12.5% higher than its base value! 300,000 To check: 1.125xIbase = (1.125) . x 1732 1948.5 A. 1125 3 100 ELEC 371 Three-phase systems 28 Industrial Power Systems Salvador Acevedo Transformers in per unit calculations With an ideal transformer + V1 - 2400 V. + V2 - 4.33 + j 2.5 ohms 2400:120 V 5 KVA High Voltage Bases Sbase1 = 5 KVA Vbase1= 2400 V Ibase1 = 5000/2400=2.083 A Zbase1= 2400/2.083=1152 Low Voltage Bases Sbase2 = 5KVA Vbase2 = 120 V I base2 = 5000/120=41.667 A Z base2 = 120/41.667=2.88 From the circuit: V1=2400 V. V2=V1/a=V1/20=120 V. In per unit: V1=1.0 p.u. V2=1.0 p.u. + 1.0 - + 1.0 - The load in per unit is: Z=(5 30)/Zbase2 =1.7361 30 p.u. The current in the circuit is: I=(1.0 0/ (1.7361 30 =0.576 -30 p.u. The current in amperes is: Primary: I1=0.576 x Ibase1= 1.2 A. Secondary: I2=0.576 x Ibase2= 24 A. ELEC 371 Three-phase systems 29 Industrial Power Systems Salvador Acevedo One line diagrams A one line diagram is a simplified representation of a multiphase-phase circuit. TRANSFORMER Transmission line TRANSFORMER GENERATOR GENERATOR Transmission line Transmission line LOAD ELEC 371 Three-phase systems 30 Industrial Power Systems Salvador Acevedo Nodal Analysis Suppose the following diagram represents the single-phase equivalent of a three-phase system z13=j2 p.u. z3=j2 p.u. z1=j1 p.u. z12=j0.5 p.u. z23=j0.5 p.u. V1= 1 p.u. V3= -j1 p.u. z2=10 p.u. Finding Norton equivalents and representing impedances as admittances: y13=-j0.5 p.u. 1 y12=-j2 p.u. 2 y1=-j1 p.u. 3 y23=-j2 p.u. y2=0.1 p.u. y3=-j0.5 p.u. I1= -j1 p.u. I3= -0.5 p.u. I1=y1 V1 + y12(V1-V2) + y13(V1-V3) 0 = y12 (V2-V1) + y2 V2 + y23(V2-V3) I3=y13(V3-V1) + y23(V3-V2) + y3 V3 In matrix form: y1 + y12 + y13 - y12 - y13 - y12 y12 + y 2 + y 23 - y 23 j 3.5 j2 0.1 j 4 j2 j 0.5 j2 ELEC 371 Three-phase systems 31 V1 - y13 - y 23 V2 y13 + y 23 + y 3 V3 j 0.5 V1 - j1 j 2 V2 0 j 3 V3 - 0.5 I1 0 I3 V1 0.77 24 solving V2 0.73 35 p. u. V3 0.71 44 Industrial Power Systems Salvador Acevedo General form of the nodal analysis The system of equations is repeated here to find a general solution technique: y1 + y12 + y13 V1 I1 - y12 - y13 y y + y + y y 12 12 2 23 23 V2 0 - y13 - y 23 y13 + y 23 + y 3 V3 I3 or Y11 Y12 Y13 V1 J1 Y Y Y 21 22 23 V2 J2 Y31 Y32 Y33 V3 J3 In general: N Yii = y ij i 1,2... N j=1 Yij = -y ij i 1,2... N ; j 1,2... N ; i j J i = I i (from current sources flowing into the node) i 1,2... N Once the voltages are found, currents and powers are easily evaluated from the circuit. We have solved one of the phases of the three-phase system (e.g. phase ‘a’). Quantities for the other two phases are shifted 120 and 240 degrees under balanced conditions. Actual quantities can be found by multiplying the per unit values by their corresponding bases. ELEC 371 Three-phase systems 32 Industrial Power Systems Salvador Acevedo